Let $g(x)=\cot(x)$. Find $g'\left(\dfrac{\pi}{4}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $\dfrac{1}{\sqrt{2}}$ (Choice C) C $-2$ (Choice D) D $0$
Solution: Let's first find $g'(x)$. Then, we can evaluate it at $x=\dfrac{\pi}{4}$. Recall that the derivative of $\cot(x)$ is $-\dfrac{1}{\sin^2(x)}$, or $-\csc^2(x)$. [Is there a way to know this without memorizing?] So $g'(x)=-\dfrac{1}{\sin^2(x)}$. Now let's plug $x={\dfrac{\pi}{4}}$ into $g'$ : $\begin{aligned} &\phantom{=}g'\left({\dfrac{\pi}{4}}\right) \\\\ &=-\dfrac{1}{\sin^2\left({\dfrac{\pi}{4}}\right)} \\\\ &=-\dfrac{1}{\left(\dfrac{\sqrt 2}{2}\right)^2} \\\\ &=-\dfrac{1}{\left(\dfrac{2}{4}\right)} \\\\ &=-2 \end{aligned}$ In conclusion, $g'\left(\dfrac{\pi}{4}\right)=-2$.